# Finding the absolute max and min of a function bounded by a domain D.

The task is to find the absolute max and absolute min of this function:

$f(x,y) = 4xy^2-x^2y^2-xy^3$

on the domain D: $D={(x,y) | x\ge0, y\ge0, x + y \le 6}$

So I get the partial derivatives of $f(x,y)$ to find the critical points

$f_x(x,y) = 4y^2-2xy^2-y^3$

$f_y(x,y) = 8xy-2x^2y-3xy^2$

Solving these for zero I get the critical point of (0,0). and f(0,0) = 0.

Then I find the extreme values at points on domain D which are (0,0),(6,0),(0,6).

In all cases the extreme values f(0,6) and f(6,0) are zero. I get the feeling I'm doing something wrong here. What are the absolute min/max values?

To find critical points (if any) in the interior of $D$, set $f_x(x,y)=y^2(4-2x-y)$ and $f_y(x,y)=xy(8-2x-3y)$ equal to $0$, and assume that $x>0,y>0,$ and $x+y<6$. Since $x,y\neq 0$ and $$y^2(4-2x-y)=0\\xy(8-2x-3y)=0,$$ it follows that $$4-2x-y=0\\8-2x-3y=0.$$ Solving this system for $y$ yields $y=2$, and back-substitution tells us that $x=1$. Indeed, $x+y=3<6$ and $x,y>0$, so this critical point is in the interior of $D$.
You've seen that $f(0,6)=f(6,0)=f(0,0)=0$, so it remains to check the boundary curves of $D$. On the lines $x=0$ and $y=0$, $f$ is constantly zero. All that's left is to check along the line $y=6-x$ for $0<x<6$. We do this by letting \begin{align}g(x) &:= f(x,6-x)\\ &= 4x(6-x)^2-x^2(6-x)^2-x(6-x)^3\\ &= (4x-x^2)(6-x)^2-(6x-x^2)(6-x)^2\\ &= -2x(6-x)^2.\end{align} Then we check for critical points of $g$ by setting $g'(x)=0$ and looking for solutions with $0<x<6$. You should find such an $x$, and then you will also need to check $(x,6-x)$ as a potential location for the global max or min of $f$.